quantum mechanics - Superposition principle forbids quantisation?

Apparently bound states in quantum mechanics require energy states to be discrete. That means energy in such systems is quantized, right? However, say that we have a superposition of energy eigenstates. Using the superposition principle, I say that their average is a possible state, since we also say that there's the same probability that we'll observe each of the two available states. Then since this average is allowed, I can take the average of that and the smaller of the original states, can't I? And if I repeat this an infinite number of times, I should get something which is as good as continuous, right? If I talk about the $n$th energy levels of some system, I won't be able to express the energy each of these allowed states in terms $n$ without making some messy equations. Isn't this a contradiction?

Answer

This can be understood from a purely classical perspective, and it's a notion which appears in the discussion of almost any discrete variable. The expectation value is not necessarily an allowed state. For instance, even with a fair die, the expectation value is 3.5 (since it's the average of the possible numbers), but you'll obviously never get 3.5.

So imagine you're dealing with a superposition of two energy eigenstates $|n_1\rangle$ and $|n_2\rangle$. The question describes a long process of averaging the energies of those states and then using that average with either $|n_1\rangle$ or $|n_2\rangle$, but that isn't even necessary to get an expectation value of some arbitrary value in between the two 'allowed' states. In your discussion of 'averages', I assume you're talking about something like this: $$\frac{1}{\sqrt{2}}|n_1\rangle+\frac{1}{\sqrt{2}}|n_2\rangle$$ But you don't need to do that whole infinity-many-averages thing. You could simply consider $$\alpha |n_1\rangle+\beta n_2\rangle$$ where $$|\alpha|^2 +|\beta|^2=1$$ and choose $\alpha$ and $\beta$ such that the expectation value is whatever you want between $|n_1\rangle$ and $|n_2\rangle$.

However, in all of this, whatever $\alpha$ and $\beta$ you consider, you will only ever observe either $|n_1\rangle$ or $|n_2\rangle$. Nothing in between.